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I know that if we want to calculate the work done on one mole of a gas quasistatically from volume, $V_i$ to volume, $V_f$ we have to calculate:

$W=-\int_{V_i}^{V_f}PdV$

But, what happens if the process is isobaric? As the pressure is constant, it would come out of the integral but then that would imply that if, for example, we have these two equations of state:

$P=\cfrac{RT}{v-b}$ and $P=\cfrac{RT}{v}\left(1-\cfrac{B}{v}\right)$

the value of work made per mole, of the corresponding substance, during a quasi-static expansion from the initial volume $v_i$ to the final volume $v_f$ will it be the same for both?

Since if it leaves the integral we can consider it as a constant $ c $ and therefore the result of the integral would be $ W = c (v_f-v_i) $ for both different equations of state.

Or the correct answers would be:

$W_1=\cfrac{RT}{v-b}(v_f-v_i)$ and $W_2=\cfrac{RT}{v}\left(1-\cfrac{B}{v}\right)(v_f-v_i)$ which causes me confusion that the random variable $ v $ exists.

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3 Answers 3

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As you said, the work done on system is defined as $$\delta W= -pdV$$

And for Isobaric process, the work done by the system will be $$\delta W = p(V_f-V_i)$$

To specify state of system uniquely, We need at least two variable. As you are taking the pressure to be constant, then one variable is already fix. The other one is, of course, the volume ( better in this case).

Now as you did, Suppose you have two different gases with different equation of state,

$$P=f(V,T) \ \ \ \mathrm{and} \ \ \ P=g(V,T)$$

Now if the pressure is fixed during isobaric process, say $P_0$, then $$P_0=f(V,T)=g(V,T)$$

So in your case during the isobaric process, You must have $$P_0=\frac{RT}{V-b}=\frac{RT}{V}\left(1-\frac{B}{V}\right)$$

So that If the volume boundary are same, then the work done in both cases will be same. But the Temperature at boundaries will be different.

$$\delta W=P_0(V_f-V_i)=\frac{RT}{V-b}(V_f-V_i)=\frac{RT}{V}\left(1-\frac{B}{V}\right)(V_f-V_i)$$

If you draw the trajectory in $V-T$ space, then the trajectory will be different but both have same value of pressure at each point. Here

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  • $\begingroup$ Thank you to pointing out. :) Beside the work is done for adiabatic. $\endgroup$ Commented Nov 26, 2020 at 8:05
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You would, of course, evaluate P at $v_i$ and $T_i$. Along the path, T would change in tandem with v so that P stays constant.

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The work done depends on the external pressure. The external pressure is independent of the equation of state for the gas. If you want to know the temperature for the initial and final equilibrium states you would need to apply the applicable equation of state.

To put it another way, you don’t need to know the equation of state to calculate the work done, but you do need to know the equation of state to determine the changes in gas properties resulting from the work done between equilibrium states.

Hope this helps.

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